Difference between revisions of "User:Aoum/Sandbox"
| Line 71: | Line 71: | ||
<span style="font-family:aops">@</span> | <span style="font-family:aops">@</span> | ||
<span style="font-family:aops">#</span> | <span style="font-family:aops">#</span> | ||
| − | <span style="font-family:aops">$</span> | + | <nowiki> <span style="font-family:aops">$</span> </nowiki> |
<span style="font-family:aops">%</span> | <span style="font-family:aops">%</span> | ||
<span style="font-family:aops">^</span> | <span style="font-family:aops">^</span> | ||
| Line 120: | Line 120: | ||
<span style="font-family:aops">—</span> <!-- em dash --> | <span style="font-family:aops">—</span> <!-- em dash --> | ||
<span style="font-family:aops">…</span> | <span style="font-family:aops">…</span> | ||
| + | <!-- | ||
| + | == Problem 1 == | ||
| + | Evaluate <math>5^2 - (10)(6) + 6^2</math>. | ||
| + | |||
| + | == Solution == | ||
| + | It's not hard to expand the expression, but we can also recognize that the given expression is an instance of the formula | ||
| + | <cmath>(a - b)^2 = a^2 - 2ab + b^2.</cmath> | ||
| + | |||
| + | We have: <math>5^2 - (10)(6) + 6^2 = 5^2 - (2)(5)(6) + 6^2 = (5-6)^2 = \boxed{1}</math>. | ||
| + | |||
| + | == Problem 2 == | ||
| + | A three digit number <math>``abc"</math> has the property that the product of <math>a</math> and <math>b</math> is equal to <math>c .</math> The digits <math>a, b,</math> and <math>c</math> are not necessarily distinct. What is the greatest possible value of the three digit number? | ||
| + | |||
| + | == Solution == | ||
| + | We want to maximize the hundreds digit above all, so let's first try setting <math>a = 9.</math> We would then like to maximize <math>b,</math> but it must satisfy the constraint <math>9b = c,</math> where <math>b</math> and <math>c</math> are digits. The only digits <math>b</math> and <math>c</math> that satisfy this equation are <math>b = 1</math> and <math>c = 9</math> or <math>b=c=0,</math> so the largest such three digit number is <math>\boxed{919}.</math> | ||
| + | |||
| + | == Problem 3 == | ||
| + | Find the smallest positive integer that is one less than a number that is a multiple of each of <math>3, 5, 7, 9,</math> and <math>11.</math> | ||
| + | |||
| + | == Solution == | ||
| + | Let <math>n</math> be the smallest positive integer that is one less than a multiple of each of <math>3, 5, 7, 9,</math> and <math>11.</math> Then <math>n+1</math> is the smallest positive integer that is a multiple of each of <math>3, 5, 7, 9,</math> and <math>11.</math> In other words, <math>n+1</math> is the least common multiple of <math>3, 5, 7, 9,</math> and <math>11.</math> | ||
| + | |||
| + | The prime factorizations of these numbers are <math>3, 5, 7, 9 = 3^2,</math> and <math>11,</math> so the lowest common multiple is <math>3^2 \cdot 5 \cdot 7 \cdot 11 = 3465.</math> Therefore, the number we seek is <math>3465 - 1 = \boxed{3464}.</math> | ||
| + | |||
| + | == Problem 4 == | ||
| + | Evaluate: <math>35^4 - 25^4</math>. | ||
| + | |||
| + | == Solution == | ||
| + | We can use difference of squares: <math>35^4-25^4 = (35^2-25^2)(35^2+25^2)</math>. (Note that every fourth power is also a square.) At this point, you can either further factor by using difference of squares on the first factor, or you can simply substitute <math>35^2 = 1225</math> and <math>25^2 = 625</math>, and evaluate: <math>(1225 - 625)(1225 + 625) = 600 \cdot 1850 = \boxed{1110000}</math>. | ||
| + | |||
| + | == Problem 5 == | ||
| + | Four different coins are used in Australia with values of <math>5, 10, 20,</math> and <math>50</math> cents. The Australian coins in Farmer Tim's pocket have a total value of <math>\$3.95,</math> and he has at least one of each coin. What is the smallest number of coins that he could have in his pocket? | ||
| + | |||
| + | == Solution == | ||
| + | Since Farmer Tim has one of each coin, <math>50+20+10+5=85</math> cents are accounted for, leaving <math>395 - 85 = 310</math> cents for the remaining coins. We can try minimizing the number of coins that add up to <math>310</math> cents by using as many large denominations as possible. | ||
| + | |||
| + | The largest denomination is <math>50</math> cents, so we can take six <math>50</math> cent coins, which add up to <math>6 \cdot 50 = 300</math> cents. We then can take one <math>10</math> cent coin. Thus, we have a combination of seven coins that up to <math>310</math> cents. | ||
| + | |||
| + | Therefore, the smallest number of coins that Farmer Tim could have is <math>4 + 7 = \boxed{11}</math>. | ||
| + | |||
| + | == Problem 6 == | ||
| + | If the integer <math>152AB1</math> is a perfect square, what is the sum of the digits of its square root? | ||
| + | |||
| + | == Solution == | ||
| + | There are two things that help us here: | ||
| + | |||
| + | First, <math>400^2 = 160000</math> which is fairly close to <math>152AB1</math>. | ||
| + | |||
| + | Second, since <math>152AB1</math> ends in a 1, its square root must end in either a 1 or a 9. | ||
| + | |||
| + | Thus, we can narrow our choices down to square roots slightly lower than 400 that end in a 1 or a 9. | ||
| + | |||
| + | Now, since <math>152AB1</math> is about 8000 smaller than 160,000, it is reasonable to guess 391 as the square root first (399 is probably a bit too close), and evaluating <math>391^2</math>, we get 152881, which fits the form <math>152AB1</math>. Therefore, the answer is <math>3 + 9 + 1 = \boxed{13}</math>. | ||
| + | |||
| + | == Problem 7 == | ||
| + | What is the largest four-digit number that is equal to the cube of the sum of its digits? | ||
| + | |||
| + | == Solution == | ||
| + | The number we seek, in particular, is a four-digit cube. Thus, we can start with the largest four-digit cube, and work our way down. | ||
| + | |||
| + | |||
| + | The largest four-digit cube is <math>21^3=9261</math>, but <math>9+2+6+1 \neq 21</math>. | ||
| + | |||
| + | |||
| + | The next lowest cube is <math>20^3=8000</math>, but <math>8+0+0+0 \neq 20</math>. | ||
| + | |||
| + | |||
| + | The next lowest cube is <math>19^3=6859</math>, but <math>6+8+5+9 \neq 19</math>. | ||
| + | |||
| + | |||
| + | The next lowest cube is <math>18^3=5832</math>, and <math>5+8+3+2=18</math>, so the answer is <math>\boxed{5832}</math>. | ||
| + | |||
| + | == Problem 8 == | ||
| + | How many different prime numbers divide evenly into <math>40^4-9^4</math>? | ||
| + | |||
| + | == Solution == | ||
| + | We replace 40 and 9 with <math>a</math> and <math>b</math> for the moment. This gives us: <math>a^4-b^4 = (a^2)^2-(b^2)^2</math>. | ||
| + | |||
| + | |||
| + | Invoking the difference of squares identity two separate times, we get: <math>a^4-b^4= (a^2+b^2)(a^2-b^2)=(a^2+b^2)(a+b)(a-b)</math>. | ||
| + | |||
| + | |||
| + | Substituting 40 and 9 back in, we have: <math>(40^2+9^2)(40+9)(40-9)</math>. The first term of this is easy to simplify if we remember the Pythagorean triple (9,40,41). We get: <math>(41^2)(49)(31)=7^2\cdot 31 \cdot 41^2</math>. | ||
| + | |||
| + | |||
| + | Now that we have the prime factorization of the given expression, we see it is divisible by the primes 7, 31, and 41, so the answer is <math>\boxed{3}</math>. | ||
| + | |||
| + | == Problem 9 == | ||
| + | Twenty-five blue and twenty-five yellow marbles are placed in a jar. Thirty-four marbles are removed from the jar and put in a second jar. What is the positive difference between the number of blue marbles in the second jar and the number of yellow marbles remaining in the first jar? | ||
| + | |||
| + | == Solution == | ||
| + | The first jar initially contains <math>25 + 25 = 50</math> marbles. After 34 marbles are moved to the second jar, the first jar has <math>50 - 34 = 16</math> marbles remaining. | ||
| + | |||
| + | Let <math>b</math> be the number of blue marbles in the first jar after the marble transfer. Then the number of yellow marbles in the first jar is <math>16 - b</math>. | ||
| + | |||
| + | There are 25 blue marbles combined between the jars, so the second jar contains <math>25 - b</math> blue marbles. Similarly, there are a total of 25 yellow marbles, so the second jar contains <math>25 - (16 - b) = b + 9</math> yellow marbles. Note that <math>(25 - b) + (b + 9) = 34</math>. | ||
| + | |||
| + | Therefore, the positive difference between the number of blue marbles in the second jar and the number of yellow marbles in the first jar is <math>(25 - b) - (16 - b) = \boxed{9}</math>. | ||
| + | |||
| + | == Problem 10 == | ||
| + | Among all fractions <math>x</math> that have a positive integer numerator and denominator and satisfy | ||
| + | <cmath>\frac{9}{11} \le x \le \frac{11}{13},</cmath> | ||
| + | which fraction has the smallest denominator? | ||
| + | |||
| + | == Solution == | ||
| + | Suppose the fraction <math>a/b</math> satisfies | ||
| + | <cmath>\frac{9}{11} \le \frac{a}{b} \le \frac{11}{13}.</cmath> | ||
| + | Then | ||
| + | <cmath>\frac{9b}{11} \le a \le \frac{11b}{13}.</cmath> | ||
| + | This inequality tells us that the positive integer <math>a</math> must lie in a certain interval. | ||
| + | |||
| + | We want to minimize <math>b</math>, so we test values of <math>b</math> starting with 1. For <math>b = 1</math>, the inequality above becomes | ||
| + | <cmath>\frac{9}{11} \le a \le \frac{11}{13}.</cmath> | ||
| + | No positive integer <math>a</math> satisfies this inequality since <math>\dfrac{11}{13}<1</math>. | ||
| + | |||
| + | For <math>b = 2</math>, the inequality above becomes | ||
| + | <cmath>\frac{18}{11} \le a \le \frac{22}{13}.</cmath> | ||
| + | No positive integer <math>a</math> satisfies this inequality because the inequality forces <math>1<a<2</math>. | ||
| + | |||
| + | We keep going, until we hit <math>b = 6</math>: | ||
| + | <cmath>\frac{54}{11} \le a \le \frac{66}{13}.</cmath> | ||
| + | The positive integer <math>a = 5</math> satisfies this inequality. Therefore, the fraction <math>x</math> that has the smallest denominator is <math>a/b = \boxed{5/6}</math>. | ||
| + | |||
| + | == Problem 11 == | ||
| + | Evaluate: <math>\dfrac{(x+y)^2 - (x-y)^2}{y}</math> for <math>x=6</math>, <math>y \not= 0</math>. | ||
| + | |||
| + | == Solution == | ||
| + | To evaluate the expression <math>(x + y)^2 - (x - y)^2</math>, we can expand it. We can also recognize that it is a difference of squares: | ||
| + | <cmath>(x + y)^2 - (x - y)^2 = [(x + y) + (x - y)][(x + y) - (x - y)] = (2x)(2y) = 4xy.</cmath> | ||
| + | |||
| + | |||
| + | Thus, | ||
| + | <cmath>\frac{(x + y)^2 - (x - y)^2}{y} = \frac{4xy}{y} = 4x = \boxed{24}.</cmath> | ||
| + | |||
| + | == Problem 12 == | ||
| + | Some perfect squares (such as 121) have a digit sum <math>(1 + 2 + 1 = 4)</math> that is equal to the square of the digit sum of their square root <math>(\sqrt{121}=11</math>, and <math>(1 + 1)^2 = 4)</math>. | ||
| + | |||
| + | What is the smallest perfect square greater than 100 that does not have this property? | ||
| + | |||
| + | == Solution == | ||
| + | We can start with <math>12^2</math> and work our way up. | ||
| + | |||
| + | We have that <math>12^2 = 144</math>, and <math>1 + 4 + 4 = 9 = (1+2)^2</math>. | ||
| + | |||
| + | Similarly, for <math>13^2 = 169</math>, we have <math>1 + 6 + 9 = 16 = (1 + 3)^2</math>. | ||
| + | |||
| + | However, for <math>14^2 = 196</math>, we have <math>1 + 9 + 6 = 16</math>, which is not equal to <math>(1+4)^2 = 25</math>. Thus, <math>\boxed{196}</math> is the smallest such square greater than 100. | ||
| + | |||
| + | (It is not a coincidence that the long multiplications for <math>11\times 11</math>, <math>12\times 12</math>, and <math>13\times 13</math> don't require carrying, but <math>14\times 14</math> does.) | ||
| + | |||
| + | == Problem 13 == | ||
| + | In the arrangement below, each number is the non-negative difference of the two numbers above it. What is the sum of the four greatest distinct possible values for <math>z</math>? | ||
| + | |||
| + | <asy> | ||
| + | unitsize(1 cm); | ||
| + | |||
| + | label("$40$", (0,0)); | ||
| + | label("$4$", (1,0)); | ||
| + | label("$18$", (2,0)); | ||
| + | label("$48$", (3,0)); | ||
| + | label("$40$", (4,0)); | ||
| + | label("$76$", (5,0)); | ||
| + | label("$z$", (6,0)); | ||
| + | label("$36$", (0.5,-0.5)); | ||
| + | label("$14$", (1.5,-0.5)); | ||
| + | label("$30$", (2.5,-0.5)); | ||
| + | label("$8$", (3.5,-0.5)); | ||
| + | label("$36$", (4.5,-0.5)); | ||
| + | label("$y$", (5.5,-0.5)); | ||
| + | label("$22$", (1,-1)); | ||
| + | label("$16$", (2,-1)); | ||
| + | label("$22$", (3,-1)); | ||
| + | label("$28$", (4,-1)); | ||
| + | label("$x$", (5,-1)); | ||
| + | label("$6$", (1.5,-1.5)); | ||
| + | label("$6$", (2.5,-1.5)); | ||
| + | label("$6$", (3.5,-1.5)); | ||
| + | label("$w$", (4.5,-1.5)); | ||
| + | label("$0$", (2,-2)); | ||
| + | label("$0$", (3,-2)); | ||
| + | label("$0$", (4,-2)); | ||
| + | </asy> | ||
| + | |||
| + | == Solution == | ||
| + | We start with the bottom of the table, and work our way up. The difference between <math>w</math> and 6 is 0, so <math>w</math> must be equal to 6. | ||
| + | |||
| + | Then the difference between <math>x</math> and 28 is 6, so <math>x</math> could be either <math>28 + 6 = 34</math> or <math>28 - 6 = 22</math>. | ||
| + | |||
| + | The difference between <math>y</math> and 36 is <math>x</math>. If <math>x = 34</math>, then <math>y</math> could be either <math>36 + 34 = 70</math> or <math>36 - 34 = 2</math>. Similarly, if <math>x = 22</math>, then <math>y</math> could be either <math>36 + 22 = 58</math> or <math>36 - 22 = 14</math>. This gives us four possible values of <math>y</math>. | ||
| + | |||
| + | The difference between <math>z</math> and 76 is <math>y</math>. For each possible value of <math>y</math>, <math>z</math> could be either <math>76 + y</math> or <math>76 - y</math>. Going through the possible values of <math>y</math>, we see that the possible values of <math>z</math> are 146, 6, 78, 74, 134, 18, 90, and 62. The largest four possible values are 146, 134, 90, and 78, and their sum is <math>146 + 134 + 90 + 78 = \boxed{448}</math>. | ||
| + | |||
| + | Alternative solution: | ||
| + | |||
| + | We have <math>w=6\pm 0,~x=28\pm w,~y=36\pm x,</math> and <math>z=76\pm y</math>. Through a series of substitutions, we obtain | ||
| + | <cmath>z = 76\pm 36\pm 28\pm 6,</cmath>which gives eight possible values for <math>z</math> (since each sign "<math>\pm</math>" can independently be chosen in two different ways). Since <math>36</math> by itself is larger than <math>28+6</math>, the four largest values of <math>z</math> are those with <math>+36</math>, regardless of the signs of <math>28</math> and <math>6</math>. That is, the four largest values of <math>z</math> are | ||
| + | \begin{align*} | ||
| + | &76+36+28+6, \\ | ||
| + | &76+36+28-6, \\ | ||
| + | &76+36-28+6, \\ | ||
| + | &76+36-28-6. | ||
| + | \end{align*}Summing, we get <math>4(76)+4(36) = \boxed{448}</math> (note that the <math>\pm28\pm6</math> terms just cancel out). | ||
| + | |||
| + | == Problem 14 == | ||
| + | A collection of nickels, dimes and pennies has an average value of 7 cents per coin. If a nickel were replaced by five pennies, the average would drop to 6 cents per coin. What is the number of dimes in the collection? | ||
| + | |||
| + | == Solution == | ||
| + | Let <math>c</math> be the number of coins and <math>v</math> the total value of the coins in cents. Our first average fact states that <math>\dfrac{v}{c}=7</math>, or <math>v=7c</math>. Once we do the exchange, we have <math>4</math> more coins and the same total value, so <math>\dfrac{v}{c+4}=6</math>, or <math>v=6(c+4)</math>. | ||
| + | |||
| + | Setting these equations equal, we have <math>7c=6(c+4)</math>, so <math>c=24</math>. Plugging this into either equation above shows <math>v=168</math> cents. | ||
| + | |||
| + | Now, let <math>p</math> be the number of pennies, <math>n</math> the number of nickels, and <math>d</math> the number of dimes. We can write the equations | ||
| + | \begin{align*} | ||
| + | d + n + p &= 24, \\ | ||
| + | 10d + 5n + p &= 168. | ||
| + | \end{align*}If we multiply the first equation by 10, and then subtract the second equation, we get <math>5n + 9p = 72</math>. Solving for <math>n</math>, we find <math>n = \dfrac{72 - 9p}{5}=\dfrac{9(8-p)}{5}</math>. | ||
| + | |||
| + | From the factorization, we see that the only <math>p</math> value which makes <math>n</math> a positive integer is <math>p=3</math>. This <math>p</math> value forces <math>n=9</math>, and thus <math>d = 24 - n - p = 24 - 3 - 9 = \boxed{12}</math>. | ||
| + | --> | ||
Revision as of 19:38, 27 June 2025
✏️ Edit this page
This Sandbox page is for experimenting with AoPS Wiki editing.
Feel free to test formatting, links, or templates here. If you're new, you may find the AoPS Wiki editing tutorial helpful.
Note: This page is cleared regularly and without warning. Please do not add offensive, copyrighted, or inappropriate content.
a b c d e f g h i j k l m n o p q r s t u v w x y z
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
0 1 2 3 4 5 6 7 8 9
! @ # <span style="font-family:aops">$</span> % ^ & * ( ) - _ = + [ ] { } | \ : ; ' " , . / < > ?
€ £ ¥ ₧ ₣ ₤ ∂ ∆
‘ ’ “ ” „ ‚ † · ˜ – — …
